3.2.17 \(\int \frac {(A+B x^2) (b x^2+c x^4)^{3/2}}{x^{17}} \, dx\)

Optimal. Leaf size=170 \[ \frac {16 c^3 \left (b x^2+c x^4\right )^{5/2} (13 b B-8 A c)}{15015 b^5 x^{10}}-\frac {8 c^2 \left (b x^2+c x^4\right )^{5/2} (13 b B-8 A c)}{3003 b^4 x^{12}}+\frac {2 c \left (b x^2+c x^4\right )^{5/2} (13 b B-8 A c)}{429 b^3 x^{14}}-\frac {\left (b x^2+c x^4\right )^{5/2} (13 b B-8 A c)}{143 b^2 x^{16}}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{13 b x^{18}} \]

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Rubi [A]  time = 0.32, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2034, 792, 658, 650} \begin {gather*} \frac {16 c^3 \left (b x^2+c x^4\right )^{5/2} (13 b B-8 A c)}{15015 b^5 x^{10}}-\frac {8 c^2 \left (b x^2+c x^4\right )^{5/2} (13 b B-8 A c)}{3003 b^4 x^{12}}+\frac {2 c \left (b x^2+c x^4\right )^{5/2} (13 b B-8 A c)}{429 b^3 x^{14}}-\frac {\left (b x^2+c x^4\right )^{5/2} (13 b B-8 A c)}{143 b^2 x^{16}}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{13 b x^{18}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^17,x]

[Out]

-(A*(b*x^2 + c*x^4)^(5/2))/(13*b*x^18) - ((13*b*B - 8*A*c)*(b*x^2 + c*x^4)^(5/2))/(143*b^2*x^16) + (2*c*(13*b*
B - 8*A*c)*(b*x^2 + c*x^4)^(5/2))/(429*b^3*x^14) - (8*c^2*(13*b*B - 8*A*c)*(b*x^2 + c*x^4)^(5/2))/(3003*b^4*x^
12) + (16*c^3*(13*b*B - 8*A*c)*(b*x^2 + c*x^4)^(5/2))/(15015*b^5*x^10)

Rule 650

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((p + 1)*(2*c*d - b*e)), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] &&
EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p + 2, 0]

Rule 658

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((m + p + 1)*(2*c*d - b*e)), x] + Dist[(c*Simplify[m + 2*p + 2])/((m + p + 1)*(2*c*d -
b*e)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c
, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[Simplify[m + 2*p + 2], 0]

Rule 792

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d*g - e*f)*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/((2*c*d - b*e)*(m + p + 1)), x] + Dist[(m*(g*(c*d - b*e)
+ c*e*f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p,
x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((L
tQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0]

Rule 2034

Int[(x_)^(m_.)*((b_.)*(x_)^(k_.) + (a_.)*(x_)^(j_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n
, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a*x^Simplify[j/n] + b*x^Simplify[k/n])^p*(c + d*x)^q, x], x, x^n], x]
/; FreeQ[{a, b, c, d, j, k, m, n, p, q}, x] &&  !IntegerQ[p] && NeQ[k, j] && IntegerQ[Simplify[j/n]] && Integ
erQ[Simplify[k/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

Rubi steps

\begin {align*} \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{17}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^9} \, dx,x,x^2\right )\\ &=-\frac {A \left (b x^2+c x^4\right )^{5/2}}{13 b x^{18}}+\frac {\left (-9 (-b B+A c)+\frac {5}{2} (-b B+2 A c)\right ) \operatorname {Subst}\left (\int \frac {\left (b x+c x^2\right )^{3/2}}{x^8} \, dx,x,x^2\right )}{13 b}\\ &=-\frac {A \left (b x^2+c x^4\right )^{5/2}}{13 b x^{18}}-\frac {(13 b B-8 A c) \left (b x^2+c x^4\right )^{5/2}}{143 b^2 x^{16}}-\frac {(3 c (13 b B-8 A c)) \operatorname {Subst}\left (\int \frac {\left (b x+c x^2\right )^{3/2}}{x^7} \, dx,x,x^2\right )}{143 b^2}\\ &=-\frac {A \left (b x^2+c x^4\right )^{5/2}}{13 b x^{18}}-\frac {(13 b B-8 A c) \left (b x^2+c x^4\right )^{5/2}}{143 b^2 x^{16}}+\frac {2 c (13 b B-8 A c) \left (b x^2+c x^4\right )^{5/2}}{429 b^3 x^{14}}+\frac {\left (4 c^2 (13 b B-8 A c)\right ) \operatorname {Subst}\left (\int \frac {\left (b x+c x^2\right )^{3/2}}{x^6} \, dx,x,x^2\right )}{429 b^3}\\ &=-\frac {A \left (b x^2+c x^4\right )^{5/2}}{13 b x^{18}}-\frac {(13 b B-8 A c) \left (b x^2+c x^4\right )^{5/2}}{143 b^2 x^{16}}+\frac {2 c (13 b B-8 A c) \left (b x^2+c x^4\right )^{5/2}}{429 b^3 x^{14}}-\frac {8 c^2 (13 b B-8 A c) \left (b x^2+c x^4\right )^{5/2}}{3003 b^4 x^{12}}-\frac {\left (8 c^3 (13 b B-8 A c)\right ) \operatorname {Subst}\left (\int \frac {\left (b x+c x^2\right )^{3/2}}{x^5} \, dx,x,x^2\right )}{3003 b^4}\\ &=-\frac {A \left (b x^2+c x^4\right )^{5/2}}{13 b x^{18}}-\frac {(13 b B-8 A c) \left (b x^2+c x^4\right )^{5/2}}{143 b^2 x^{16}}+\frac {2 c (13 b B-8 A c) \left (b x^2+c x^4\right )^{5/2}}{429 b^3 x^{14}}-\frac {8 c^2 (13 b B-8 A c) \left (b x^2+c x^4\right )^{5/2}}{3003 b^4 x^{12}}+\frac {16 c^3 (13 b B-8 A c) \left (b x^2+c x^4\right )^{5/2}}{15015 b^5 x^{10}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 95, normalized size = 0.56 \begin {gather*} \frac {\sqrt {x^2 \left (b+c x^2\right )} \left (\left (\frac {c x^3}{b}+x\right )^2 \left (105 b^3-70 b^2 c x^2+40 b c^2 x^4-16 c^3 x^6\right ) (8 A c-13 b B)-1155 A b^2 \left (b+c x^2\right )^2\right )}{15015 b^3 x^{14}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^17,x]

[Out]

(Sqrt[x^2*(b + c*x^2)]*(-1155*A*b^2*(b + c*x^2)^2 + (-13*b*B + 8*A*c)*(x + (c*x^3)/b)^2*(105*b^3 - 70*b^2*c*x^
2 + 40*b*c^2*x^4 - 16*c^3*x^6)))/(15015*b^3*x^14)

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IntegrateAlgebraic [A]  time = 0.52, size = 162, normalized size = 0.95 \begin {gather*} \frac {\sqrt {b x^2+c x^4} \left (-1155 A b^6-1470 A b^5 c x^2-35 A b^4 c^2 x^4+40 A b^3 c^3 x^6-48 A b^2 c^4 x^8+64 A b c^5 x^{10}-128 A c^6 x^{12}-1365 b^6 B x^2-1820 b^5 B c x^4-65 b^4 B c^2 x^6+78 b^3 B c^3 x^8-104 b^2 B c^4 x^{10}+208 b B c^5 x^{12}\right )}{15015 b^5 x^{14}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^17,x]

[Out]

(Sqrt[b*x^2 + c*x^4]*(-1155*A*b^6 - 1365*b^6*B*x^2 - 1470*A*b^5*c*x^2 - 1820*b^5*B*c*x^4 - 35*A*b^4*c^2*x^4 -
65*b^4*B*c^2*x^6 + 40*A*b^3*c^3*x^6 + 78*b^3*B*c^3*x^8 - 48*A*b^2*c^4*x^8 - 104*b^2*B*c^4*x^10 + 64*A*b*c^5*x^
10 + 208*b*B*c^5*x^12 - 128*A*c^6*x^12))/(15015*b^5*x^14)

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fricas [A]  time = 0.61, size = 157, normalized size = 0.92 \begin {gather*} \frac {{\left (16 \, {\left (13 \, B b c^{5} - 8 \, A c^{6}\right )} x^{12} - 8 \, {\left (13 \, B b^{2} c^{4} - 8 \, A b c^{5}\right )} x^{10} + 6 \, {\left (13 \, B b^{3} c^{3} - 8 \, A b^{2} c^{4}\right )} x^{8} - 1155 \, A b^{6} - 5 \, {\left (13 \, B b^{4} c^{2} - 8 \, A b^{3} c^{3}\right )} x^{6} - 35 \, {\left (52 \, B b^{5} c + A b^{4} c^{2}\right )} x^{4} - 105 \, {\left (13 \, B b^{6} + 14 \, A b^{5} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{15015 \, b^{5} x^{14}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^17,x, algorithm="fricas")

[Out]

1/15015*(16*(13*B*b*c^5 - 8*A*c^6)*x^12 - 8*(13*B*b^2*c^4 - 8*A*b*c^5)*x^10 + 6*(13*B*b^3*c^3 - 8*A*b^2*c^4)*x
^8 - 1155*A*b^6 - 5*(13*B*b^4*c^2 - 8*A*b^3*c^3)*x^6 - 35*(52*B*b^5*c + A*b^4*c^2)*x^4 - 105*(13*B*b^6 + 14*A*
b^5*c)*x^2)*sqrt(c*x^4 + b*x^2)/(b^5*x^14)

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giac [B]  time = 4.41, size = 550, normalized size = 3.24 \begin {gather*} \frac {32 \, {\left (15015 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{18} B c^{\frac {11}{2}} \mathrm {sgn}\relax (x) - 3003 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{16} B b c^{\frac {11}{2}} \mathrm {sgn}\relax (x) + 48048 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{16} A c^{\frac {13}{2}} \mathrm {sgn}\relax (x) - 6006 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{14} B b^{2} c^{\frac {11}{2}} \mathrm {sgn}\relax (x) + 96096 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{14} A b c^{\frac {13}{2}} \mathrm {sgn}\relax (x) - 28314 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{12} B b^{3} c^{\frac {11}{2}} \mathrm {sgn}\relax (x) + 109824 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{12} A b^{2} c^{\frac {13}{2}} \mathrm {sgn}\relax (x) + 13728 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{10} B b^{4} c^{\frac {11}{2}} \mathrm {sgn}\relax (x) + 37752 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{10} A b^{3} c^{\frac {13}{2}} \mathrm {sgn}\relax (x) + 5720 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} B b^{5} c^{\frac {11}{2}} \mathrm {sgn}\relax (x) + 5720 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} A b^{4} c^{\frac {13}{2}} \mathrm {sgn}\relax (x) + 3718 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} B b^{6} c^{\frac {11}{2}} \mathrm {sgn}\relax (x) - 2288 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} A b^{5} c^{\frac {13}{2}} \mathrm {sgn}\relax (x) - 1014 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} B b^{7} c^{\frac {11}{2}} \mathrm {sgn}\relax (x) + 624 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} A b^{6} c^{\frac {13}{2}} \mathrm {sgn}\relax (x) + 169 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} B b^{8} c^{\frac {11}{2}} \mathrm {sgn}\relax (x) - 104 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} A b^{7} c^{\frac {13}{2}} \mathrm {sgn}\relax (x) - 13 \, B b^{9} c^{\frac {11}{2}} \mathrm {sgn}\relax (x) + 8 \, A b^{8} c^{\frac {13}{2}} \mathrm {sgn}\relax (x)\right )}}{15015 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{13}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^17,x, algorithm="giac")

[Out]

32/15015*(15015*(sqrt(c)*x - sqrt(c*x^2 + b))^18*B*c^(11/2)*sgn(x) - 3003*(sqrt(c)*x - sqrt(c*x^2 + b))^16*B*b
*c^(11/2)*sgn(x) + 48048*(sqrt(c)*x - sqrt(c*x^2 + b))^16*A*c^(13/2)*sgn(x) - 6006*(sqrt(c)*x - sqrt(c*x^2 + b
))^14*B*b^2*c^(11/2)*sgn(x) + 96096*(sqrt(c)*x - sqrt(c*x^2 + b))^14*A*b*c^(13/2)*sgn(x) - 28314*(sqrt(c)*x -
sqrt(c*x^2 + b))^12*B*b^3*c^(11/2)*sgn(x) + 109824*(sqrt(c)*x - sqrt(c*x^2 + b))^12*A*b^2*c^(13/2)*sgn(x) + 13
728*(sqrt(c)*x - sqrt(c*x^2 + b))^10*B*b^4*c^(11/2)*sgn(x) + 37752*(sqrt(c)*x - sqrt(c*x^2 + b))^10*A*b^3*c^(1
3/2)*sgn(x) + 5720*(sqrt(c)*x - sqrt(c*x^2 + b))^8*B*b^5*c^(11/2)*sgn(x) + 5720*(sqrt(c)*x - sqrt(c*x^2 + b))^
8*A*b^4*c^(13/2)*sgn(x) + 3718*(sqrt(c)*x - sqrt(c*x^2 + b))^6*B*b^6*c^(11/2)*sgn(x) - 2288*(sqrt(c)*x - sqrt(
c*x^2 + b))^6*A*b^5*c^(13/2)*sgn(x) - 1014*(sqrt(c)*x - sqrt(c*x^2 + b))^4*B*b^7*c^(11/2)*sgn(x) + 624*(sqrt(c
)*x - sqrt(c*x^2 + b))^4*A*b^6*c^(13/2)*sgn(x) + 169*(sqrt(c)*x - sqrt(c*x^2 + b))^2*B*b^8*c^(11/2)*sgn(x) - 1
04*(sqrt(c)*x - sqrt(c*x^2 + b))^2*A*b^7*c^(13/2)*sgn(x) - 13*B*b^9*c^(11/2)*sgn(x) + 8*A*b^8*c^(13/2)*sgn(x))
/((sqrt(c)*x - sqrt(c*x^2 + b))^2 - b)^13

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maple [A]  time = 0.04, size = 118, normalized size = 0.69 \begin {gather*} -\frac {\left (c \,x^{2}+b \right ) \left (128 A \,c^{4} x^{8}-208 B b \,c^{3} x^{8}-320 A b \,c^{3} x^{6}+520 B \,b^{2} c^{2} x^{6}+560 A \,b^{2} c^{2} x^{4}-910 B \,b^{3} c \,x^{4}-840 A \,b^{3} c \,x^{2}+1365 B \,b^{4} x^{2}+1155 A \,b^{4}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{15015 b^{5} x^{16}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^17,x)

[Out]

-1/15015*(c*x^2+b)*(128*A*c^4*x^8-208*B*b*c^3*x^8-320*A*b*c^3*x^6+520*B*b^2*c^2*x^6+560*A*b^2*c^2*x^4-910*B*b^
3*c*x^4-840*A*b^3*c*x^2+1365*B*b^4*x^2+1155*A*b^4)*(c*x^4+b*x^2)^(3/2)/b^5/x^16

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maxima [B]  time = 1.62, size = 337, normalized size = 1.98 \begin {gather*} \frac {1}{9240} \, B {\left (\frac {128 \, \sqrt {c x^{4} + b x^{2}} c^{5}}{b^{4} x^{2}} - \frac {64 \, \sqrt {c x^{4} + b x^{2}} c^{4}}{b^{3} x^{4}} + \frac {48 \, \sqrt {c x^{4} + b x^{2}} c^{3}}{b^{2} x^{6}} - \frac {40 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{b x^{8}} + \frac {35 \, \sqrt {c x^{4} + b x^{2}} c}{x^{10}} + \frac {315 \, \sqrt {c x^{4} + b x^{2}} b}{x^{12}} - \frac {1155 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{14}}\right )} - \frac {1}{30030} \, A {\left (\frac {256 \, \sqrt {c x^{4} + b x^{2}} c^{6}}{b^{5} x^{2}} - \frac {128 \, \sqrt {c x^{4} + b x^{2}} c^{5}}{b^{4} x^{4}} + \frac {96 \, \sqrt {c x^{4} + b x^{2}} c^{4}}{b^{3} x^{6}} - \frac {80 \, \sqrt {c x^{4} + b x^{2}} c^{3}}{b^{2} x^{8}} + \frac {70 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{b x^{10}} - \frac {63 \, \sqrt {c x^{4} + b x^{2}} c}{x^{12}} - \frac {693 \, \sqrt {c x^{4} + b x^{2}} b}{x^{14}} + \frac {3003 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{16}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^17,x, algorithm="maxima")

[Out]

1/9240*B*(128*sqrt(c*x^4 + b*x^2)*c^5/(b^4*x^2) - 64*sqrt(c*x^4 + b*x^2)*c^4/(b^3*x^4) + 48*sqrt(c*x^4 + b*x^2
)*c^3/(b^2*x^6) - 40*sqrt(c*x^4 + b*x^2)*c^2/(b*x^8) + 35*sqrt(c*x^4 + b*x^2)*c/x^10 + 315*sqrt(c*x^4 + b*x^2)
*b/x^12 - 1155*(c*x^4 + b*x^2)^(3/2)/x^14) - 1/30030*A*(256*sqrt(c*x^4 + b*x^2)*c^6/(b^5*x^2) - 128*sqrt(c*x^4
+ b*x^2)*c^5/(b^4*x^4) + 96*sqrt(c*x^4 + b*x^2)*c^4/(b^3*x^6) - 80*sqrt(c*x^4 + b*x^2)*c^3/(b^2*x^8) + 70*sqr
t(c*x^4 + b*x^2)*c^2/(b*x^10) - 63*sqrt(c*x^4 + b*x^2)*c/x^12 - 693*sqrt(c*x^4 + b*x^2)*b/x^14 + 3003*(c*x^4 +
b*x^2)^(3/2)/x^16)

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mupad [B]  time = 2.50, size = 306, normalized size = 1.80 \begin {gather*} \frac {8\,A\,c^3\,\sqrt {c\,x^4+b\,x^2}}{3003\,b^2\,x^8}-\frac {14\,A\,c\,\sqrt {c\,x^4+b\,x^2}}{143\,x^{12}}-\frac {B\,b\,\sqrt {c\,x^4+b\,x^2}}{11\,x^{12}}-\frac {4\,B\,c\,\sqrt {c\,x^4+b\,x^2}}{33\,x^{10}}-\frac {A\,c^2\,\sqrt {c\,x^4+b\,x^2}}{429\,b\,x^{10}}-\frac {A\,b\,\sqrt {c\,x^4+b\,x^2}}{13\,x^{14}}-\frac {16\,A\,c^4\,\sqrt {c\,x^4+b\,x^2}}{5005\,b^3\,x^6}+\frac {64\,A\,c^5\,\sqrt {c\,x^4+b\,x^2}}{15015\,b^4\,x^4}-\frac {128\,A\,c^6\,\sqrt {c\,x^4+b\,x^2}}{15015\,b^5\,x^2}-\frac {B\,c^2\,\sqrt {c\,x^4+b\,x^2}}{231\,b\,x^8}+\frac {2\,B\,c^3\,\sqrt {c\,x^4+b\,x^2}}{385\,b^2\,x^6}-\frac {8\,B\,c^4\,\sqrt {c\,x^4+b\,x^2}}{1155\,b^3\,x^4}+\frac {16\,B\,c^5\,\sqrt {c\,x^4+b\,x^2}}{1155\,b^4\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^17,x)

[Out]

(8*A*c^3*(b*x^2 + c*x^4)^(1/2))/(3003*b^2*x^8) - (14*A*c*(b*x^2 + c*x^4)^(1/2))/(143*x^12) - (B*b*(b*x^2 + c*x
^4)^(1/2))/(11*x^12) - (4*B*c*(b*x^2 + c*x^4)^(1/2))/(33*x^10) - (A*c^2*(b*x^2 + c*x^4)^(1/2))/(429*b*x^10) -
(A*b*(b*x^2 + c*x^4)^(1/2))/(13*x^14) - (16*A*c^4*(b*x^2 + c*x^4)^(1/2))/(5005*b^3*x^6) + (64*A*c^5*(b*x^2 + c
*x^4)^(1/2))/(15015*b^4*x^4) - (128*A*c^6*(b*x^2 + c*x^4)^(1/2))/(15015*b^5*x^2) - (B*c^2*(b*x^2 + c*x^4)^(1/2
))/(231*b*x^8) + (2*B*c^3*(b*x^2 + c*x^4)^(1/2))/(385*b^2*x^6) - (8*B*c^4*(b*x^2 + c*x^4)^(1/2))/(1155*b^3*x^4
) + (16*B*c^5*(b*x^2 + c*x^4)^(1/2))/(1155*b^4*x^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}} \left (A + B x^{2}\right )}{x^{17}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(c*x**4+b*x**2)**(3/2)/x**17,x)

[Out]

Integral((x**2*(b + c*x**2))**(3/2)*(A + B*x**2)/x**17, x)

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